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 A209083 Largest number of the form C(n,x) + C(n,y) + C(n,z) where x + y + z = n. 1
 3, 3, 5, 9, 14, 25, 45, 77, 141, 261, 505, 935, 1849, 3445, 6865, 12885, 25741, 48637, 97241, 184775, 369513, 705453, 1410865, 2704179, 5408313, 10400625, 20801201, 40116627, 80233201, 155117549, 310235041, 601080421, 1202160781, 2333606253, 4667212441 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS lim{n->infinity} a(n+1)/a(n)=2. Subset of A034703. From an idea of Michael B. Porter [michael_b_porter(AT)yahoo.com]. For n > 6, it appears that the solution is always x = n mod 2, y = z = floor(n/2). - T. D. Noe, Mar 05 2012 LINKS Paolo P. Lava, Table of n, a(n) for n = 0..360 EXAMPLE For n=5 [x,y,z] can be [0,0,5], [0,1,4], [0,2,3], [1,1,3] and [1,2,2]. C(5,0) + C(5,0) + C(5,5) = 1+1+1 = 3. C(5,0) + C(5,1) + C(5,4) = 1+5+5 = 11. C(5,0) + C(5,2) + C(5,3) = 1+10+10 =21. C(5,1) + C(5,1) + C(5,3) = 5+5+10 = 20. C(5,1) + C(5,2) + C(5,2) = 5+10+10 = 25. Therefore 25 is in the sequence. MAPLE with(numtheory); P:=proc(i) local c, m, n, s, v; v:=array[1..3]; for n from 3 to i do   s:=0; v[1]:=0; v[2]:=0; v[3]:=n;   while v[1]<=floor(n/3) do     while v[2]<=floor((n-v[1])/2) do       c:=0;       for m from 1 to 3 do c:=c+binomial(n, v[m]); od;       if c>s then s:=c; fi;       v[2]:=v[2]+1; v[3]:=v[3]-1;     od;     v[1]:=v[1]+1; v[2]:=v[1]; v[3]:=n-v[1]-v[2];   od;   print(s); od; end: P(1000); MATHEMATICA Table[Maximize[{Binomial[n, a] + Binomial[n, b] + Binomial[n, c], a + b + c == n, a >= 0, b >= 0, c >= 0, a <= n, b <= n, c <= n}, {a, b, c}, Integers][[1]], {n, 0, 30}] (* T. D. Noe, Mar 05 2012 *) PROG (PARI) A209083(n)={local(a, b, c, s); s=-1; for(a=0, n, for(b=0, n-a, c=n-a-b; s=max(s, binomial(n, a)+binomial(n, b)+binomial(n, c)))); s} CROSSREFS Cf. A034703. Sequence in context: A159284 A078028 A104220 * A137202 A146926 A000198 Adjacent sequences:  A209080 A209081 A209082 * A209084 A209085 A209086 KEYWORD nonn AUTHOR Paolo P. Lava, Mar 05 2012 STATUS approved

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